How to Unwrap a Modal class object in Swift 3 - ios

Question

Model Class:
class CountryModel: NSObject {
var name:NSString!
var countryId:NSString!
init(name: NSString, countryId: NSString) {
self.name = name
self.countryId = countryId
}
}
ViewController:
var nameArr = [CountryModel]() // Model Class object
func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
var cell = UITableViewCell()
cell = tableView.dequeueReusableCell(withIdentifier: "cell")!
let arr = nameArr[indexPath.row] // How to do if let here
let str:String? = "\(arr.countryId) \(arr.name)"
if let sstr = str{
cell.textLabel?.text = sstr
}
return cell
}
How should one unwrap this because output is an optional, if I try to unwrap nameArr[indexPath.row] gives an error initializer for conditional binding must have optional type, not "country modal"
It works fine I am not concatenating arr.countryId with arr.name

Answer 1

This works fine
var nameArr = [CountryModal]()
nameArr.append(CountryModal.init(name: "harshal", countryId: "india"))
nameArr.append(CountryModal.init(name: "james", countryId: "australia"))
let arr = nameArr[0]
let str:String? = "\(arr.countryId!) \(arr.name!)"
if let sstr = str{
print(sstr)
}
let arr2 = nameArr[1]
let str2:String? = "\(arr2.countryId!) \(arr2.name!)"
if let sstr2 = str2{
print(sstr2)
}

Answer 2

You can try this library https://github.com/T-Pham/NoOptionalInterpolation. It does exactly that
import NoOptionalInterpolation
let n: Int? = 1
let t: String? = nil
let s: String? = "string1"
let o: String?? = "string2"
let i = "\(n) \(t) \(s) \(o)"
print(i) // 1 string1 string2
NoOptionalInterpolation gets rid of "Optional(...)" and "nil" in Swift's string interpolation. This library ensures that the text you set never ever includes that annoying additional "Optional(...)". You can also revert to the default behaviour when needed.
Also, please note that this does not affect the print function. Hence, print(o) (as opposed to print("(o)"), o as in the example above) would still print out Optional(Optional("string2"))

Answer 3

Avoid forced unwrapping as much as possible. i.e. using '!'. Whenever you see !, try and think of it as code smell.
For json parsing you would not know for sure if the values for countryId and name exists or not. So, it makes sense to keep them as optionals and gracefully try to unwrap them later.
class CountryModel {
var name: String?
var countryId: String?
init(name: String?, countryId: String?) {
self.name = name
self.countryId = countryId
}
}
var nameArr = [CountryModel]()
func tableView(_ tableView: UITableView, cellForRowAt indexPath: IndexPath) -> UITableViewCell {
guard let cell = tableView.dequeueReusableCell(withIdentifier: "cell") else {
fatalError("cell should always be initialised")
}
var str: String?
let arr = nameArr[indexPath.row] // How to do if let here
if let countryId = arr.countryId, let name = arr.name {
str = "\(countryId) \(name)"
}
if let unwrappedString = str {
cell.textLabel?.text = unwrappedString
}
return cell
}
Bonus:
You could simplify your json parsing a lot using the Object Mapper library. It simplifies a lot of your parsing logic.